Proving an identity

**xhazel06x** · March 26th 2009, 06:21 PM

Establish the identity:

(cotθ / 1- tanθ) + (tanθ / 1 – cotθ) = 1+ tanθ + cotθ

**Grandad** · March 26th 2009, 10:39 PM

Hello xhazel06x

Originally Posted by xhazel06x

Establish the identity:

(cotθ / 1- tanθ) + (tanθ / 1 – cotθ) = 1+ tanθ + cotθ

If we let $\tan\theta = t$ , then

$\frac{\cot\theta}{1-\tan\theta}+\frac{\tan\theta}{1-\cot\theta}= \frac{\frac{1}{t}}{1-t}+\frac{t}{1-\frac{1}{t}}$

$=\frac{\frac{1}{t}}{1-t}+\frac{t^2}{t-1}$

$=\frac{\frac{1}{t}}{1-t}-\frac{t^2}{1-t}$

$=\frac{\frac{1}{t}-t^2}{1-t}$

$=\frac{\frac{1}{t}(1-t^3)}{1-t}$

$=\frac{\frac{1}{t}(1-t)(1+t+t^2)}{1-t}$

$=\frac{1}{t} + 1 + t$

$= 1 + \tan\theta+\cot\theta$

Grandad

**Soroban** · March 27th 2009, 02:24 AM

Hello, xhazel06x!

Another approach . . . longer than Grandad's

Establish the identity: . $\frac{\cot\theta}{1-\tan\theta} + \frac{\tan\theta}{1-\cot\theta} \:=\: 1+ \tan\theta + \cot\theta$

The left side is: . $\frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}} + \frac{\dfrac{\sin\theta}{\cos\theta}} {1 - \dfrac{\cos\theta}{\sin\theta}}$

Multiply the fractions by $\frac{\sin\theta\cos\theta}{\sin\theta\cos\theta}$

$\frac{\sin\theta\cos\theta\left(\dfrac{\cos\theta} {\sin\theta}\right)}{\sin\theta\cos\theta\left(1 - \dfrac{\sin\theta}{\cos\theta}\right)} + \frac{\sin\theta\cos\theta\left(\dfrac{\sin\theta} {\cos\theta}\right)} {\sin\theta\cos\theta\left(1 - \dfrac{\cos\theta}{\sin\theta}\right)}$ . $= \;\frac{\cos^2\!\theta}{\sin\theta\cos\theta - \sin^2\!\theta} + \frac{\sin^2\!\theta}{\sin\theta\cos\theta - \cos^2\!\theta}$

. . $=\;\frac{\cos^2\!\theta}{\sin\theta(\cos\theta-\sin\theta)} + \frac{\sin^2\!\theta}{-\cos\theta(\cos\theta - \sin\theta)}$ . $= \;{\color{blue}\frac{\cos\theta}{\cos\theta}}\cdot \frac{\cos^2\!\theta}{\sin\theta(\cos\theta - \sin\theta)} \;- \;{\color{blue}\frac{\sin\theta}{\sin\theta}}\cdot \frac{\sin^2\!\theta}{\cos\theta(\cos\theta - \sin\theta)}$

. . $= \;\frac{\cos^3\!\theta}{\sin\theta\cos\theta(\cos\ theta-\sin\theta)} - \frac{\sin^3\!\theta}{\sin\theta\cos\theta(\cos\th eta - \sin\theta)}$ . $= \;\frac{\cos^3\!\theta - \sin^3\theta}{\sin\theta\cos\theta(\cos\theta - \sin\theta)}$

. . $= \;\frac{(\cos\theta-\sin\theta)(\cos^2\!\theta + \cos\theta\sin\theta + \sin^2\!\theta)}{\sin\theta\cos\theta(\cos\theta - \sin\theta)}$ . $= \;\frac{\cos^2\!\theta + \cos\theta\sin\theta + \sin^2\!\theta}{\sin\theta\cos\theta}$

. . $=\; \frac{\cos^2\!\theta}{\sin\theta\cos\theta} + \frac{\cos\theta\sin\theta}{\sin\theta\cos\theta} + \frac{\sin^2\!\theta}{\sin\theta\cos\theta}$ . $=\; \frac{\cos\theta}{\sin\theta} + 1 + \frac{\sin\theta}{\cos\theta}$

. . $=\;\cot\theta + 1 + \tan\theta$

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